¿CUAL ES EL PERIMETRO DE UN RECTANGULO SABIENDO QUE SU ANCHO ES 3/4 DEL LARGO Y BEL LARGO MIDE 2,55M?

I need help in math. I think it is x > -6

Alborosie

Answer:

C. x > 6

Step-by-step explanation:

Solve for x by simplifying both sides of the inequality, then isolating the variable.

3 0

3 years ago

Read 2 more answers

What is 69 over 25 as a terminating decimal

frutty [35]

The answer is 2.76 because you would do 69 divided by 25

3 0

4 years ago

Read 2 more answers

What is the equation of a line perpendicular to

Valentin [98]

Since the given line

$y = \frac{ 1}{4} x - 3$

has slope
$= \frac{1}{4}$

The equation of the line perpendicular to it must have a slope which is the negative reciprocal of ,
$\frac{1}{4}$
The slope of the perpendicular line
$= \frac{ - 1}{ \frac{1}{4} } = - 4$
Using the slope intercept form,

$y = mx + c$
We substitute
$- 4$
This implies that,
$y = - 4x + c$
Since

$(-2,4)$
lies on this line , it must satisfy its equation.

That is

$4 = - 4( - 2) + c$

This implies that

$4 = 8 + c$
$4 - 8 = c$
$c = - 4$

The line therefore has equation,

$y = - 4x - 4$

3 0

3 years ago

A golf ball is hit with a club from the ground on a flat surface. The height of the ball, h(t), in feet, t seconds after it was

Arlecino [84]

Answer:

The correct numbers are;

1) 0

2) 25

3) 1.25

4) 2.5

When the ball is hit at 0 seconds, it has a height of 0 feet. The ball's height increases until it reaches its maximum height of 25 feet after 1.25 seconds. The ball's height then decreases until it reaches the ground 2.5 seconds after it was hit

Step-by-step explanation:

The given equation for the height of the ball, h(t), in feet, is given as follows

h(t) = 40·t - 16·t²

Therefore;

1) When the ball is hit at 0 seconds, it has a height of h(0) = 40×0 - 16×0² = 0 feet

2) The time to reach maximum height by the ball is found by taking the derivative (slope) of the function for the height of the ball and noting that at maximum height, the derivative (slope) = 0

Therefore;

h'(t) = 40 - 2 × 16·t = 0

40 - 32·t = 0

t = 40/32 = 5/4 seconds = 1.25 seconds

3) The maximum height is then found from the value of the function at t = 5/4 seconds as follows;

h(5/4) = 40×5/4 - 16×(5/4)² = 50 - 25 = 25 feet

4) The time in which the ball reaches the ground where h(t) = 0 is given from the formula for the height of the gulf ball as follows;

h(t) = 0 = 40·t - 16·t²

40·t - 16·t² = 0

(40 - 16·t)·t = 0

Therefore;

t = 0 seconds or t = 40/16 = 2.5 seconds

(40 - 16·t) = 0/t = 0

t = 40/16 = 2.5 seconds

Therefore, the filling the blanks with the correct numbers gives;

When the ball is hit at 0 seconds, it has a height of 0 feet. The ball's height increases until it reaches its maximum height of 25 feet after 1.25 seconds. The ball's height then decreases until it reaches the ground 2.5 seconds after it was hit

4 0

3 years ago

Find the area of the region that is inside r=3cos(theta) and outside r=2-cos(theta). Sketch the curves.

raketka [301]

Answer:

3√3

Step-by-step explanation:

r = 3 cos θ

r = 2 - cos θ

First, find the intersections.

3 cos θ = 2 - cos θ

4 cos θ = 2

cos θ = 1/2

θ = -π/3, π/3

We want the area inside the first curve and outside the second curve. So R = 3 cos θ and r = 2 - cos θ, such that R > r.

Now that we have the limits, we can integrate.

A = ∫ ½ (R² - r²) dθ

A = ∫ ½ ((3 cos θ)² - (2 - cos θ)²) dθ

A = ∫ ½ (9 cos² θ - (4 - 4 cos θ + cos² θ)) dθ

A = ∫ ½ (9 cos² θ - 4 + 4 cos θ - cos² θ) dθ

A = ∫ ½ (8 cos² θ + 4 cos θ - 4) dθ

A = ∫ (4 cos² θ + 2 cos θ - 2) dθ

Using power reduction formula:

A = ∫ (2 + 2 cos(2θ) + 2 cos θ - 2) dθ

A = ∫ (2 cos(2θ) + 2 cos θ) dθ

Integrating:

A = (sin (2θ) + 2 sin θ) |-π/3 to π/3

A = (sin (2π/3) + 2 sin(π/3)) - (sin (-2π/3) + 2 sin(-π/3))

A = (½√3 + √3) - (-½√3 - √3)

A = 1.5√3 - (-1.5√3)

A = 3√3

The area inside of r = 3 cos θ and outside of r = 2 - cos θ is 3√3.

The graph of the curves is:

desmos.com/calculator/541zniwefe

5 0

3 years ago